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    Normal Distribution

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    Pages C H A P T E R 6 The Normal Distribution Objectives Outline After completing this chapter‚ you should be able to 1 2 3 Identify distributions as symmetric or skewed. 4 Find probabilities for a normally distributed variable by transforming it into a standard normal variable. Introduction 6–1 Normal Distributions Identify the properties of a normal distribution. Find the area under the standard normal distribution‚ given various z values. 5 Find

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    Normal Distribution

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    Normal Distribution:- A continuous random variable X is a normal distribution with the parameters mean and variance then the probability function can be written as f(x) =   - < x < ‚ - < μ < ‚  σ > 0. When σ2 = 1‚ μ = 0 is called as standard normal. Normal distribution problems and solutions – Formulas: X < μ = 0.5 – Z X > μ = 0.5 + Z X = μ = 0.5 where‚ μ = mean σ = standard deviation X = normal random variable Normal Distribution Problems and Solutions – Example

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    Normal Distribution

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    Normal Distribution It is important because of Central Limit Theorem (CTL)‚ the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal. Normal P.D.F Now we want to find c This integral has been proved that it cannot have close form solution. However‚ someone gives an idea that looks stupid but actually very brilliant by multiply two of them. reminds the function of circle which we can replace them to polar coordinate Thus Mean

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    Normal Distribution

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    NORMAL DISTRIBUTION 1. Find the distribution: a. b. c. d. e. f. following probabilities‚ the random variable Z has standard normal P (0< Z < 1.43) P (0.11 < Z < 1.98) P (-0.39 < Z < 1.22) P (Z < 0.92) P (Z > -1.78) P (Z < -2.08) 2. Determine the areas under the standard normal curve between –z and +z: ♦ z = 0.5 ♦ z = 2.0 Find the two values of z in standard normal distribution so that: P(-z < Z < +z) = 0.84 3. At a university‚ the average height of 500 students of a course is 1.70 m; the standard

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    Normal Distribution

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    standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data 34‚ 38‚ 22‚ 21‚ 29‚ 37‚ 40‚ 41‚ 22‚ 20‚ 49‚ 47‚ 20‚ 31‚ 34‚ 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot? 45 40 35 30 25 20 15 10 5 0 0 5 10 15 20 Solution: Weak positive linear

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    Normal Distribution

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    Normal Distribution Normal distribution is a statistics‚ which have been widely applied of all mathematical concepts‚ among large number of statisticians. Abraham de Moivre‚ an 18th century statistician and consultant to gamblers‚ noticed that as the number of events (N) increased‚ the distribution approached‚ forming a very smooth curve. He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to

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    Normal Distribution

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    decimal places) 2. Find the value of z if the area under a Standard Normal curve a) to the right of z is 0.3632; b) to the left of z is 0.1131; c) between 0 and z‚ with z > 0‚ is 0.4838; d) between -z and z‚ with z > 0‚ is 0.9500. Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table) b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table) c ) the area between 0 to z is 0.4838‚ z = 2.14 d) the area to the

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    Normal Distribution

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    found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3)‚ we have z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2 From the area between z =±2 is 2(0.4772) = 0.9554 Therefore the probability that a measurement selected at random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values. σ = 3 0.95 σ = 1 0.95 X 19 25 31 -2 0 +2 Normal curve showing

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    the five-number summary and the mean and standard deviation of the data. C. Describe the distribution of the data‚ citing both the plots and the summary statistics found in questions 1 and 2. AP Statistics Exam Review Topic II: Normal Distribution [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] FREE RESPONSE A set of 2‚000 measurements had a symmetric‚ mound-shaped distribution. The mean is 5.3 and the standard deviation is 0.7. Determine an interval that contains

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    2 Normal Distribution

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    Business Statistics MGSC-372 Review Normal Distribution The Normal Distribution aka The Gaussian Distribution The Normal Distribution y 1 f ( x)  e 2  1  x      2   2 x Areas under the Normal Distribution curve -3 -2 -  68% 95% 99.7% + +2 +3 X = N( ‚ 2 ) Determining Normal Probabilities Since each pair of values for  and  represents a different distribution‚ there are an infinite number of possible normal distributions. The number of statistical tables

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